Upperbound approximation to the sum of Euler's totient function

I am currently working on a solution to a problem related to the density of finite coprime sets. I believe that I have found a solution to this problem - though it can only be expressed in terms of the sum of Euler's totient function:

$$\Phi(N) = \sum_{i=1}^N \phi(i)$$

Although $\Phi(N)$ is well studied, many of the expressions for this term are not easy to work with from an analytical perspective (i.e. they depend on the Möbius function, or on recursive formulations....)

I am wondering if anyone knows of a closed-form expression which provides an upper-bound approximation to $\Phi(N)$.

The best one that I can come up with is somewhat obvious:

$$S_N = \sum_{i=1}^N \phi(i) \leq \sum_{i=1}^N i = \frac{N(N+1)}{2}$$

A better version of this bound exploits the fact that $\phi(i) \leq \frac{i}{2}$, for even numbers (proposed by Hagen von Eitzen in the comments). This leads to:

\begin{align} S_N &= \sum_{i=1}^N \phi(i) \\ &\leq \sum_{k=1}^{\lfloor \frac{N}{2} \rfloor} \phi(2k) + \sum_{k=1}^{\lfloor \frac{N+1}{2} \rfloor} \phi(2k-1)\\ &\leq \sum_{k=1}^{\lfloor \frac{N}{2} \rfloor} \phi(k) + \sum_{k=1}^{\lfloor \frac{N+1}{2} \rfloor} \phi(2k-1)\\ &\leq \sum_{k=1}^{\lfloor \frac{N}{2} \rfloor} k + \sum_{k=1}^{\lfloor \frac{N+1}{2} \rfloor} 2k-1\\ &=\frac{1}{2} \Bigg(\bigg \lfloor \frac{N}{2} \bigg \rfloor \Bigg) \Bigg(\bigg \lfloor \frac{N}{2} \bigg \rfloor + 1\Bigg) + \Bigg(\bigg \lfloor \frac{N+1}{2} \bigg \rfloor \Bigg)^2\\ \end{align}

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Well, have the $i$ are even and have $\phi(i)\le \frac i2$. That should give you an improvement to an upper bound $\approx \frac38n^2$ isntead of $\frac12 n^2$, but that'ts not much –  Hagen von Eitzen Oct 15 '13 at 5:53

The proof of the asymptotic formula

$$\Phi(n) = \frac{3}{\pi^2} n^2 + O(n\log n)$$

can be modified to give a simple upper bound that has the right asymptotic character (but is not very accurate). We'll follow the proof in Chandrasekharan's Introduction to Analytic Number Theory.

By interpreting $\Phi(n)$ as the number of lattice points with relatively prime coordinates in or on the triangle $0 < y \leq x \leq n$, Chandrasekharan obtains the formula

$$\Phi(n) = \frac{\Psi(n)+1}{2},$$

where

$$\Psi(n) = \sum_{1 \leq d \leq n} \mu(d) \left\lfloor \frac{n}{d} \right\rfloor^2.$$

If we write

$$\left\lfloor \frac{n}{d} \right\rfloor = \frac{n}{d} - \left\{\frac{n}{d}\right\},$$

then

\begin{align} \Psi(n) &= \sum_{1 \leq d \leq n} \mu(d)\left(\frac{n}{d} - \left\{\frac{n}{d}\right\}\right)^2 \\ &= n^2 \sum_{1 \leq d \leq n} \frac{\mu(d)}{d^2} - 2n \sum_{1 \leq d \leq n} \frac{\mu(d)}{d}\left\{\frac{n}{d}\right\} + \sum_{1 \leq d \leq n} \mu(d) \left\{\frac{n}{d}\right\}^2. \end{align}

Now

\begin{align} -2n \sum_{1 \leq d \leq n} \frac{\mu(d)}{d}\left\{\frac{n}{d}\right\} &< 2n \sum_{1 \leq d \leq n} \frac{1}{d} \\ &< 2n \left(1+\int_1^n \frac{du}{u}\right) \\ &= 2n(1+\log n) \end{align}

and

$$\sum_{1 \leq d \leq n} \mu(d) \left\{\frac{n}{d}\right\}^2 < \sum_{1 \leq d \leq n} 1 = n,$$

giving us

$$\Psi(n) < n^2 \sum_{1 \leq d \leq n} \frac{\mu(d)}{d^2} + 2n\log n + 3n.$$

For the sum we get

\begin{align} \sum_{1 \leq d \leq n} \frac{\mu(d)}{d^2} &= \sum_{1 \leq d \leq \infty} \frac{\mu(d)}{d^2} - \sum_{n+1 \leq d \leq \infty} \frac{\mu(d)}{d^2} \\ &= \frac{6}{\pi^2} - \sum_{n+1 \leq d \leq \infty} \frac{\mu(d)}{d^2} \\ &< \frac{6}{\pi^2} + \sum_{n+1 \leq d \leq \infty} \frac{1}{d^2} \\ &< \frac{6}{\pi^2} + \int_n^\infty \frac{du}{u^2} \\ &= \frac{6}{\pi^2} + \frac{1}{n}, \end{align}

so that, in total,

$$\Psi(n) < \frac{6}{\pi^2} n^2 + 2n\log n + 4n$$

and hence

$$\Phi(n) < \frac{3}{\pi^2} n^2 + n\log n + 2n + \frac{1}{2}.$$

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